I was given an exercise to prove that the group of $n \times n$ unitriangular matrices with entries in a commutative ring is solvable.
My attempt:
Let $(K, +, \bullet)$ be any commutative ring. We will refer to the $+$ operation as "addition" (s.t. $0$ denotes the zero element) and the $\bullet$ operation as "multiplication" (s.t. $1$ denotes the trivial element and by assumption $1 \neq 0$). $K$, in general, will denote the group $(K, +)$ under addition, and $K^\times$ shall denote the group $(K\setminus \{0\}, \bullet)$.
Let us, in fact, begin with a group more general than the group of $n \times n$ unitriangular matrices:
$$ B_n = \left \{\begin{bmatrix} a_1 & b_{1,2} & \cdots & b_{1,n} \\ & \ddots & & \vdots \\ & & a_{n-1} & b_{n-1, n} \\ & & & a_n \end{bmatrix} \mid b_{i,j} \in K, a_i \in K^\times\right\}.$$
Consider the surjective homomorphism $\varphi: B_n \to (K^\times)^n$ defined by
$$\varphi\left(\begin{bmatrix} a_1 & b_{1,2} & \cdots & b_{1,n} \\ & \ddots & & \vdots \\ & & a_{n-1} & b_{n-1, n} \\ & & & a_n \end{bmatrix}\right) = (a_1,....,a_n).$$
This is certainly a valid homomorphism because we know that if $\mathbf{A} = [a]_n$ and $\mathbf{B} = [b]_n$ are both upper triangular matrices of order $n$, then the diagonal elements of $\mathbf{C} = \mathbf{AB}$ is given by $$c_{ii} = a_{ii}b_{ii} \ \forall i \in \{1, \ldots, n\}$$ and this property should hold true even when all the matrix entries belong to the group $K^\times$.
It is clear that
$$M_0 = \mathrm{Ker}(\varphi) = \left \{\begin{bmatrix} 1 & b_{1,2} & \cdots & b_{1,n} \\ & \ddots & & \vdots \\ & & 1 & b_{n-1, n} \\ & & & 1 \end{bmatrix} \mid b_{i,j} \in K, a_i \in K^\times\right\}$$ which is precisely the group of $n \times n$ unitriangular matrices. Then $B_n/M_0 \cong (K^\times)^n$ is abelian.
We can now find a normal subsequence of $M_0$ by considering matrices with more and more partial diagonal layers of $0$s in the upper triangle. Let us define $N_k$ as the subgroup of the group $M_0$, containing specifically those matrices in $M_0$ whose first $k$ upper partial diagonals contain $0$. Or rather, those matrices in which the entries $b_{i,j}$ for $1 \leq j - i \leq k$ are zero.
It is now simple to show that $N_{k-1}$ is a normal subgroup of $N_k$ for any $k \in \{1, \ldots, n\}$. We consider the surjective homomorphism
$$\varphi_k: N_{k-1} \to K^{n-k} \\ (b_{i, j})_{j - i =k} \mapsto (b_{i, j+k})$$
which basically copies out the first upper partial diagonal that doesn't necessarily contain all $0$s. We can infer from here that $\mathrm{Ker}(\varphi_k) = N_k$, which implies any $N_k$ is normal in $N_{k-1}$. Moreover, $N_{k-1}/N_{k}$ is abelian since $K$ is commutative under addition and multiplication. This gives us a subnormal sequence with abelian factors:
$$B_n \unrhd M_0 = N_0 \unrhd N_1 \unrhd N_2 \cdots \unrhd N_{n-2} \unrhd N_{n-1} = \{1\}.$$
This explicitly proves that $B_n$, and in turn $M_0$, is solvable.
Questions:
Will this proof remain the same in case $K$ is a commutative ring rather than a field? Does the (non-)invertibility of non-zero elements of $K$ have any effect on the proof? Is matrix multiplication any different in case the entries are from a commutative ring rather than a field?
Is the fact that $K$ is commutative under addition and multiplication sufficient to prove that $N_{k-1}/N_k$ is abelian?
(I) The fact that $K$ is a commutative ring with $1$ and not a field makes no difference to your argument.
To see this note that each element of the unitriangular group is of the form $I+N$ where $N$ is nilpotent. The group inverse of this element is then $I-N+N^2-\dots\pm N^{n-1}$: that is it depends only on the addition and multiplication in $K$: we never need to inverses of elements of $K$.
(II) A slightly longer answer is this: what you are calculating seems to be the lower central series of the unitriangular group, and so you are proving something stronger than mere solubility. Let me explain.
Let $\mathfrak{n}$ denote the set of strictly upper triangular matrices; these are of course nilpotent. It is an easy calculation that $\mathfrak{n}^k$ is the set of matrices which only have non-zero entries $x_{ij}$ when $j-i\geqslant k$. It is also straightforward to see that $\mathfrak{n}^r\mathfrak{n}^s=\mathfrak{n}^{r+s}$.
Let $G_k$ denote the set of matrices $I+\mathfrak{n}^k$. This contains the identity, is closed under multiplication by our remark on the powers of $\mathfrak{n}$, and closed under inverses by the nilpotence property. So $G_k$ is a group, and $G_1$ is the unitriangular group itself.
Now let us calculate $[G_1,G_k]$. This is generated by all $(I+N)^{-1}(I+M)^{-1}(I+N)(I+M)$ where $N\in\mathfrak{n}$, $M\in\mathfrak{n}^k$. Working modulo $\mathfrak{n}^{k+1}$ we have that this commutator is $$ (I+N)^{-1}(I-M)(I+N)(I+M)=(I+N)^{-1}(I+N-MN+NM)=(I+N)^{-1}(I+N), $$ using the fact that $NM, MN\in\mathfrak{n}^{k+1}$.
That is, we have proved that $[G_1,G_k]\leqslant G_{k+1}$, and so we have shown that the lower central series terminates at $G_n=\{I\}$. (I think the $G_k$ are the terms of the lower central series, but that needs proof.)
(III) To prove mere solubility I think (see my comment) that it is easier to proceed by induction, and consider the homomorphism $\begin{bmatrix}1 & x\\0 &A\end{bmatrix}\mapsto A$. The kernel of this is the set of all $\begin{bmatrix}1 & x\\0 &I\end{bmatrix}$, an abelian group isomorphic the additive group of all $x\in K^{n-1}$. No more needs to be said.